Calculating Heat Loss from Room with Refrigerator and Heater

[oliverjames]

A room contains 25 kg of air at 100KPa and 10 degree C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running) and 1-KW electric resistance heater. During a cold winter day, it is observed that both the refrigerator and the resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room during that day is:

I`m having trouble solving this . can anyone please help?

cheers
james

 

[russ_watters]

It's a bit of a trick question, but when you see the trick, it is rediculously easy (that should be a hint...).

 

[oliverjames]

ok . if i omit the room criteria and if i`m taking only the refrigerator data the heat loss to the room would be 1250 Kw. correct me if i was wrong. Thanks . James

 

[russ_watters]

Yep, it really is that easy. Nice! Conservation of energy says that if you're at an equilibrium, the inputs and outputs must be equal.

 

[sardar]

I would approach this problem by first breaking it down into smaller parts and using the laws of thermodynamics to analyze the situation.

Firstly, we know that the room contains 25 kg of air at 100 KPa and 10 degrees C. This gives us the initial conditions for the air in the room.

Next, we need to consider the two heat sources in the room - the 250-W refrigerator and the 1-KW electric resistance heater. These devices consume energy and convert it into heat, which is then transferred to the air in the room through convection.

According to the First Law of Thermodynamics, the rate of heat transfer into the room must be equal to the rate of heat loss from the room in order for the temperature to remain constant. This means that the heat being generated by the refrigerator and heater must be equal to the heat being lost from the room.

Using the equation Q = mCΔT, where Q is the heat transfer, m is the mass of the air, C is the specific heat capacity of air, and ΔT is the change in temperature, we can calculate the amount of heat being generated by each device.

For the refrigerator, Q = (250 W) x (1 day) = 21,600,000 J
For the heater, Q = (1000 W) x (1 day) = 86,400,000 J

Now, we can use the Second Law of Thermodynamics to determine the rate of heat loss from the room. This law states that heat will always flow from a warmer object to a cooler object, so the air in the room will lose heat to the colder outside environment.

We can use the equation Q = UAΔT, where Q is the heat transfer, U is the overall heat transfer coefficient, A is the surface area of the room, and ΔT is the temperature difference between the inside and outside of the room.

Assuming the room is well-insulated, we can estimate U to be around 0.1 W/m2K. The surface area of the room will vary, but for simplicity, let's assume it is 100 m2. The temperature difference between the inside and outside of the room can be taken as 10 degrees C (since the temperature inside the room remains constant).

Plugging in these values, we get Q = (0.1 W/m2K) x